The following solution was sent by `kevin295'.

The segments $CD$ and $EF$ are parallel.

PROOF:

$ABDC$ is a cyclic quadrilateral, therefore
angle $CDB$ + angle $CAB=180^{\circ}$ (1)

Angle $CAB$ and angle $BAE$ are adjacent angles on a straight line so
angle $BAE$ + angle $CAB = 180^{\circ}$ (2)

From (1) and (2) we get
angle $CDB$ = angle $BAE$ (3)

Because $ABFE$ is a cyclic quadrilateral,
angle $BAE$ + angle $BFE = 180^{\circ}$ (4)

From (3) and (4) we get
angle $CDB$ + angle $BFE = 180^{\circ}$

so $CD$ is parallel to $EF$.